Hello,
Can you please help with this:
If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?
OA: [spoiler]29/128[/spoiler]
When I saw the word at-least 5 times, I used the following approach:
P(Heads at-least 5 times) = 1 - P(No heads at least 5 times)
= 1 - (1/2.1/2.1/2.1/2.1/2)
= 1 - (1/2)^5
= 1 - 1/32
= 31/32
However, it doesn't seem correct. Can you please assist?
Thanks a lot,
Sri
If Jesse flips a coin seven times in a row, probablity that
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Hi Sri,gmattesttaker2 wrote:Hello,
Can you please help with this:
If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?
OA: [spoiler]29/128[/spoiler]
When I saw the word at-least 5 times, I used the following approach:
P(Heads at-least 5 times) = 1 - P(No heads at least 5 times)
= 1 - (1/2.1/2.1/2.1/2.1/2)
= 1 - (1/2)^5
= 1 - 1/32
= 31/32
However, it doesn't seem correct. Can you please assist?
Thanks a lot,
Sri
Here is an approach you could follow:
We are looking for a head atleast 5 times.
P(Atleast 5 times head) = P(Exactly 5 times head)+P(Exactly 6 times head) + P(Exactly 7 times head)
Let us calculate P(Exactly 5 times head) :
Probability of Getting 5 heads = 1/2 * 1/2* 1/2 * 1/2 * 1/2 = 1/2^5
Probability of Getting 2 tails = 1/2 * 1/2 = 1/4
Out of the 7 occurrences, We need only 5 occurrences of head. Hence we can obtain them in 7C5 ways.
Therefore, probability = 7C5 * 1/2^5 * 1/4 = 21/128
Applying the same logic Let us calculate P(Exactly 6 times head) :
Probability of Getting 6 heads = 1/2 * 1/2* 1/2 * 1/2 * 1/2 * 1/2= 1/2^6
Probability of Getting 1 tails = 1/2
Out of the 7 occurrences, We need only 6 occurrences of head. Hence we can obtain them in 7C6 ways.
Therefore, probability = 7C6 * 1/2^6 * 1/2 = 7/128
P(Exactly 7 times head) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/2^7
Adding all these, we get 21/128 + 7/128 + 1/128 = [spoiler]29/128[/spoiler]
Hope this helps!!
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don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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Hello KVCP,kvcpk wrote:Hi Sri,gmattesttaker2 wrote:Hello,
Can you please help with this:
If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?
OA: [spoiler]29/128[/spoiler]
When I saw the word at-least 5 times, I used the following approach:
P(Heads at-least 5 times) = 1 - P(No heads at least 5 times)
= 1 - (1/2.1/2.1/2.1/2.1/2)
= 1 - (1/2)^5
= 1 - 1/32
= 31/32
However, it doesn't seem correct. Can you please assist?
Thanks a lot,
Sri
Here is an approach you could follow:
We are looking for a head atleast 5 times.
P(Atleast 5 times head) = P(Exactly 5 times head)+P(Exactly 6 times head) + P(Exactly 7 times head)
Let us calculate P(Exactly 5 times head) :
Probability of Getting 5 heads = 1/2 * 1/2* 1/2 * 1/2 * 1/2 = 1/2^5
Probability of Getting 2 tails = 1/2 * 1/2 = 1/4
Out of the 7 occurrences, We need only 5 occurrences of head. Hence we can obtain them in 7C5 ways.
Therefore, probability = 7C5 * 1/2^5 * 1/4 = 21/128
Applying the same logic Let us calculate P(Exactly 6 times head) :
Probability of Getting 6 heads = 1/2 * 1/2* 1/2 * 1/2 * 1/2 * 1/2= 1/2^6
Probability of Getting 1 tails = 1/2
Out of the 7 occurrences, We need only 6 occurrences of head. Hence we can obtain them in 7C6 ways.
Therefore, probability = 7C6 * 1/2^6 * 1/2 = 7/128
P(Exactly 7 times head) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/2^7
Adding all these, we get 21/128 + 7/128 + 1/128 = [spoiler]29/128[/spoiler]
Hope this helps!!
Thank you very much for your detailed explanation.
Best Regards,
Sri
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Probability = Favorable outcomes / Total outcomes
With 7 coins tossed, Total Outcomes = 2^7 = 128
Favorable out comes are as follows
1) Exactly 5 heads out of 7 coins = 7C5 = 21
2) Exactly 6 heads out of 7 coins = 7C6 = 7
3) Exactly 7 heads out of 7 coins = 7C7 = 1
Probability = (21+7+1)/128 = 29/128 ANSWER
With 7 coins tossed, Total Outcomes = 2^7 = 128
Favorable out comes are as follows
1) Exactly 5 heads out of 7 coins = 7C5 = 21
2) Exactly 6 heads out of 7 coins = 7C6 = 7
3) Exactly 7 heads out of 7 coins = 7C7 = 1
Probability = (21+7+1)/128 = 29/128 ANSWER
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We have 3 scenarios:gmattesttaker2 wrote:Hello,
Can you please help with this:
If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?
Scenario 1: 5 heads and 2 tails (e.g., H-H-H-H-H-T-T)
P(H-H-H-H-H-T-T) = (1/2)^7 = 1/128
Number of ways to arrange H-H-H-H-H-T-T = 7C5 = 7!/(5! x 2!) = (7 x 6)/2 = 21
Thus, the probability of this scenario is 1/128 x 21 = 21/128.
Scenario 2: 6 heads and 1 tail (e.g., H-H-H-H-H-H-T)
P(H-H-H-H-H-H-T) = (1/2)^7 = 1/128
Number of ways to arrange H-H-H-H-H-H-T = 7C6 = 7!/6! = 7
Thus the probability of this scenario is 1/128 x 7 = 7/128.
Scenario 3: all 7 heads (i.e., H-H-H-H-H-H-H)
P(H-H-H-H-H-H-H) = (1/2)^7 = 1/128
So, P(at least 5 heads) = 21/128 + 7/128 + 1/128 = [spoiler]29/128[/spoiler]
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